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Difference between revisions of "2015 AMC 10A Problems/Problem 20"

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==Problem==
 
==Problem==
  
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A square with positive integer side lengths in <math>\text{cm}</math> has area <math>A</math> <math>\text{cm}^2</math> and perimeter <math>P</math> <math>\text{cm}</math>. Which of the following numbers cannot equal <math>A+P</math>?
+
A rectangle with positive integer side lengths in <math>\text{cm}</math> has area <math>A</math> <math>\text{cm}^2</math> and perimeter <math>P</math> <math>\text{cm}</math>. Which of the following numbers cannot equal <math>A+P</math>?
  
 
<math> \textbf{(A) }100\qquad\textbf{(B) }102\qquad\textbf{(C) }104\qquad\textbf{(D) }106\qquad\textbf{(E) }108 </math>
 
<math> \textbf{(A) }100\qquad\textbf{(B) }102\qquad\textbf{(C) }104\qquad\textbf{(D) }106\qquad\textbf{(E) }108 </math>
  
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NOTE: The original version of this problem was incorrect, as it was not stated that the side lengths must be positive integers, so <math>A+P</math> could equal any of the answer choices. This problem was thrown out, and all contestants received full credit for it (whether they answered correctly, incorrectly, or left it blank). The problem above was modified to have an answer; as follows is the original problem:
+
==Solution==
  
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A rectangle has area <math>A</math> <math>\text{cm}^2</math> and perimeter <math>P</math> <math>\text{cm}</math>, where <math>A</math> and <math>P</math> are positive integers. Which of the following numbers cannot equal <math>A+P</math>?
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Let the rectangle's length be <math>a</math> and its width be <math>b</math>. Its area is <math>ab</math> and the perimeter is <math>2a+2b</math>.
  
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==Solution==
+
Then <math>A + P = ab + 2a + 2b</math>. Factoring, we have <math>(a + 2)(b + 2) - 4</math>.
 +
 
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The only one of the answer choices that cannot be expressed in this form is <math>102</math>, as <math>102 + 4</math> is twice a prime. There would then be no way to express <math>106</math> as <math>(a + 2)(b + 2)</math>, keeping <math>a</math> and <math>b</math> as positive integers.
  
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Let the rectangle's length and width be <math>a</math> and <math>b</math>. Its area is <math>ab</math> and the perimeter is <math>2(a + b)</math>.
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Our answer is then <math>\boxed{B}</math>.
  
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Then <math>A + P = ab + 2a + 2b</math>. Factoring, this is <math>(a + 2)(b + 2) - 4</math>.
+
Note: The original problem only stated that <math>A</math> and <math>P</math> were positive integers, not the side lengths themselves. This rendered the problem unsolvable, and so the AMC awarded everyone 6 points on this problem. This wiki has the corrected version of the problem so that the 2015 AMC 10A test can be used for practice.
  
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Looking at the answer choices, only <math>102</math> cannot be written this way, because then either <math>a</math> or <math>b</math> would be <math>0</math>.
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==Video Solution==
 +
https://youtu.be/RLo-e2On6Ac
  
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So the answer is <math>\boxed{\textbf{(B) }102}</math>.
+
~savannahsolver
  
 
==See Also==
 
==See Also==

Latest revision as of 19:16, 17 November 2020

Problem

A rectangle with positive integer side lengths in $\text{cm}$ has area $A$ $\text{cm}^2$ and perimeter $P$ $\text{cm}$. Which of the following numbers cannot equal $A+P$?

$\textbf{(A) }100\qquad\textbf{(B) }102\qquad\textbf{(C) }104\qquad\textbf{(D) }106\qquad\textbf{(E) }108$

Solution

Let the rectangle's length be $a$ and its width be $b$. Its area is $ab$ and the perimeter is $2a+2b$.

Then $A + P = ab + 2a + 2b$. Factoring, we have $(a + 2)(b + 2) - 4$.

The only one of the answer choices that cannot be expressed in this form is $102$, as $102 + 4$ is twice a prime. There would then be no way to express $106$ as $(a + 2)(b + 2)$, keeping $a$ and $b$ as positive integers.

Our answer is then $\boxed{B}$.

Note: The original problem only stated that $A$ and $P$ were positive integers, not the side lengths themselves. This rendered the problem unsolvable, and so the AMC awarded everyone 6 points on this problem. This wiki has the corrected version of the problem so that the 2015 AMC 10A test can be used for practice.

Video Solution

https://youtu.be/RLo-e2On6Ac

~savannahsolver

See Also

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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